If we denote the Cartesian coordinate of a -dimensional Euclidean space by a -vector , then a -ball centered at the origin with radius is the set of points that satisfy
Taking this definition, a 1-ball is the line segment , which its “volume” is and has no surface area. A 2-ball is a circle with radius , which has the “surface” area of and “volume” of .
Let us denote the volume of an -ball with radius by , and consider that -ball is defined by
Therefore, we can find recursively:
Proportionality of Volume
Firstly, we prove is proportional to by induction:
Consider , . If we have , then
Formula for Volume
If we denote for ,
valid for if we specially define , and which
where the Gamma function is defined as
with the property that , and for positive integers . But we can also consider without resolving to the Gamma function. Consider that
It is easy to see that , , , and
Therefore we have
Indeed, we can also prove by induction that
And the surface area of the -ball is obtained by .
Below is an alternative derivation by Greg Huber (1982). “Gamma function derivation of n-sphere volumes”. Am. Math. Monthly 89(5):301–302 (PDF)
for the reason that .
Denote the surface of -ball by , we can write the volume as the integration of surface area as the radius expands from zero:
or the volume as the integration of all the points in the ball:
Consider the density integration on an infinitely large -ball, where the density at point is with defined by
Using the first integral we have (assumed radius 1)
But using the second integral we have
Because they shall be equal, we have
We can also verify that .
Properties of n-ball
If we consider the -ball with unit diameter instead of unit radius, its volume is . We can see that, due to the exponential nature in the denominator,
Therefore, the higher the dimension, the denser we can pack the balls. I first learned this result from Professor Raymond Yeung.